Answer
$2[(3x-1)^{2}+(1-x)^{5}]\cdot[6(3x-1)-5(1-x)^{4}]$
Work Step by Step
Let $u(x)=x^{2},\quad v(x)=(3x-1)^{2}+(1-x^{5})^{2}.$
Then
$\displaystyle \frac{d}{dx}[f(x)]=\frac{du}{dv}\frac{dv}{dx}$
$v(x)$ is a sum of two composite functions
$v(x)=r(x)+s(x)$
$r(x)=(3x-1)^{2}, $
$r^{\prime}(x)=2(3x-1)\cdot(3x-1)^{\prime}=6(3x-1)$
$s(x)=(1-x)^{5}$
$s^{\prime}(x)=5(1-x)^{4}\cdot(1-x)^{\prime}=-5(1-x)^{4}$
$\displaystyle \frac{dv}{dx}=6(3x-1)-5(1-x)^{4}$
$\displaystyle \frac{du}{dv}=2v=2[(3x-1)^{2}+(1-x)^{5}]$
$\displaystyle \frac{d}{dx}[f(x)]=\frac{du}{dv}\frac{dv}{dx}$
$=2[(3x-1)^{2}+(1-x)^{5}]\cdot[6(3x-1)-5(1-x)^{4}]$
